Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int mx = 1e5 + 10;
long long int binpow(long long int base, long long int exp) {
  long long int res = 1;
  while (1) {
    if (exp & 1) res *= base;
    exp >>= 1;
    if (!exp) break;
    base *= base;
  }
  return res;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> v;
  string ans = "NO\n";
  for (long long int i = 1; ((i * (i + 1)) / 2) <= n; ++i) {
    long long int k = (i * (i + 1)) / 2;
    v.push_back(k);
  }
  for (long long int i = 1; ((i * (i + 1)) / 2) <= n; ++i) {
    long long int k = (i * (i + 1)) / 2;
    long long int baki = n - k;
    if (binary_search(v.begin(), v.end(), baki)) {
      ans = "YES\n";
      break;
    }
  }
  cout << ans;
  return 0;
}